Problem: $g(n) = 3n-6$ $h(n) = -n^{2}+3n+g(n)$ $ g(h(-7)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(-7)$ . Then we'll know what to plug into the outer function. $h(-7) = -(-7)^{2}+(3)(-7)+g(-7)$ To solve for the value of $h$ , we need to solve for the value of $g(-7)$ $g(-7) = (3)(-7)-6$ $g(-7) = -27$ That means $h(-7) = -(-7)^{2}+(3)(-7)-27$ $h(-7) = -97$ Now we know that $h(-7) = -97$ . Let's solve for $g(h(-7))$ , which is $g(-97)$ $g(-97) = (3)(-97)-6$ $g(-97) = -297$